A ball is thrown upward with a velocity of 20 m/s from the ground.
a. In how much time, will it reach the top of its path? 
b. It falls back on the ground, what is the total time of its trip? 
c. How high will it rise?

Ans. (2 s, 4 s, 20 m )


Given data:


Initial Velocity  = `\vec v_i` = 20 m s⁻¹

Final Velocity  = `\vec v_f` = 0 m s⁻¹ 

Gravitational acceleration = g = - 10 m s⁻² (-ve for upward motion)


To Find:


Time to reach maximum height = t = ? 

Total flight time = ?

The maximum height reached = h = ? m


Solution:  



Here, we are using the 1st equation of motion

vf = vi + at

or

t  = `\frac {vf -vi}{g}`

by putting values

t  = `\frac {0 m s⁻¹ - 20 m s⁻¹}{- 10 m s⁻²}`

by simplifying we get

t  = 2 s---------An.1
Thus, the time (upwards) to reach maximum height is 2s.


As we know in vertical motion under gravitational full, the body move with uniform acceleration so, the time taken to reach the maximum height from the ground will the same as the body returns to the ground. So, in this case

Total flight Time of the Cricket ball  = time in the upwards motion + time in the downwards motion

As we have
Time in the upwards motion = Time in the downwards motion = 2 s

Total flight time of the ball  = 2 s + 2 s = 4 s. ---------------Ans.2



Now to calculate the maximum height we are using the 2nd equation of motion for veridical motion where the replacement of (a with g) and (S with h) so,

h = vi t  + `\frac{1}{2}`gt²

by putting vales

h = 2o ms⁻¹ï½˜ 2 s + `\frac{1}{2}`(-10 m s⁻²) (2 s)²

h = 40 m - 20 m

h = 20 m

So, the maximum height that the tennis ball will reach is 20 meters. 



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